Sec_2.4 - 2.4 Exact Equations Let M ( x, y ) and N ( x, y )...

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Unformatted text preview: 2.4 Exact Equations Let M ( x, y ) and N ( x, y ) be continuous and have continuous first partial derivatives in a rectangular region R defined by a < x < b and c < y < d . Then a necessary and sufficient condition that M ( x, y ) dx + N ( x, y ) dy be an exact differential is xM N = . y x Ex. Show that the equation is exact. 2 a. ( 2 x - 5 y ) dx + ( -5 x + 3 y ) dy = 0 M ( x, y ) = 2 x - 5 y and N ( x, y ) = -5 x + 3 y 2 . Then Theorem 2.2 Criterion for an Exact Differential x ( -5 x + 3 y 2 ) x ( 2x - 5 y ) = -5, = -5 . y x It is an exact equation. 2 3 Let's look at the equation if x - 5 xy + y = c , then x ( x 2 - 5 xy + y 3 ) x ( x 2 - 5 xy + y 3 ) = -5 x + 3 y 2 . Clearly, = 2 x - 5 y and y x If the function f ( x, y ) = c is a solution to the exact differential equation, then x f ( x, y ) f ( x, y ) = M ( x, y ) and = N ( x, y ) . x y Method of Proof & Solution x f ( x, y ) = M ( x, y ) f ( x, y ) = M ( x, y ) dx + g ( y ) x f = M ( x, y ) dx + g ( y ) = N ( x, y ) y y g x( y ) = N ( x, y ) - x M ( x, y ) dx y ( y) 0 g = x ( x, y ) - N M ( x, y ) dx 0 = x y N - M ( x, y ) dx 0 = x y x N M - = 0. x y 2 Ex. Solve 2 xydx + ( x - 1) dy = 0 . To solve this kind of differential equation, it is necessary to see whether the equation is exact or not. xM N M ( x, y ) = 2 xy = 2 x, N ( x , y ) = x 2 - 1 = 2x . y x Clearly, it is exact. Then xf = 2 xy f ( x, y ) = x 2 y + g ( y ) x f = 2 y + g ( y ) x 2 + g ( y ) = x 2 - 1 x = y y g ( y ) = -1 g ( y ) = - y f ( x, y ) = x 2 y - y = c. e2 y dx 2 2 y dy Ex. Solve - y cos ( xy ) + xe - x cos ( xy ) + 2 y = 0 . xf = e2 y - y cos ( xy ) f ( x, y ) = 2 y - y cos ( xy ) e dx x f = xe 2 y - sin ( xy ) + g ( y ) xf = 2 y - sin ( xy ) + g ( y ) xe y y g ( y ) = 2 y g ( y ) = y2 = 2 xe 2 y - x cos ( xy ) + g ( y ) = 2 xe 2 y - x cos ( xy ) + 2 y f ( x, y ) = xe 2 y - sin ( xy ) + y 2 = c. Ex. Solve the initial problem ( x ) sin ( x ) - xy 2 + ( 1 - x 2 ) = 0, y ( 0 ) = 2 . cos dx y dy Solution: xf 1 = cos ( x ) sin ( x ) - xy 2 f ( x, y ) = cos 2 ( x ) - x 2 y 2 g ( y ) - + x 2 f 1 = cos 2 ( x ) - x 2 y 2 g ( y ) - + y y 2 = - x2 y + g ( y ) = y ( 1 - x2 ) g ( y) = y g ( y) = 1 2 y 2 1 1 f ( x, y ) = cos 2 ( x ) - x 2 y 2 y 2 = c. - + 2 2 Ex. Solve ( x + y ) dx + x ln ( x ) dy = 0 , using u ( x, y ) = Solution: y 1 dx Rewrite ( x + y ) dx + x ln ( x ) dy = 0 to + + ln ( x ) dy = 0 . x xf y = 1 + f ( x, y ) = x + y ln ( x ) + g ( y ) x x f = ln ( x ) + g ( y ) = ln ( x ) y g ( y) = 0 g ( y) = c f ( x, y ) = x + y ln ( x ) + c = 0. Homework: 1~43 odd. 1 , on ( 0,h ) . x ...
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This note was uploaded on 04/01/2008 for the course MATH 2214 taught by Professor Edesturler during the Fall '06 term at Virginia Tech.

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